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Represent the number of items bought by Juma, Tom, and Mary using a 3x2 matrix. Represent the prices of items bought using a 2x1 matrix. Use the matrices to determine the total expenditure incurred by

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Here's the solution to question 22.

Part 22

Part a(i)

Represent the number of items bought by Juma, Tom, and Mary using a 3×23 \times 2 matrix.

Juma bought 3 dozens of Maize flour, which is 3×12=363 \times 12 = 36 packets.
The items are Maize flour (packets) and Sugar (kg).

Quantity Matrix (Q)=[362045183825]\text{Quantity Matrix (Q)} = \begin{bmatrix} 36 & 20 \\ 45 & 18 \\ 38 & 25 \end{bmatrix}

Where rows represent Juma, Tom, Mary respectively, and columns represent Maize flour, Sugar respectively.

Part a(ii)

Represent the prices of items bought using a 2×12 \times 1 matrix.

The price of a packet of maize flour is Ksh 170.
The price of a kilogram of sugar is Ksh 210.

Price Matrix (P)=[170210]\text{Price Matrix (P)} = \begin{bmatrix} 170 \\ 210 \end{bmatrix}

Where rows represent Maize flour, Sugar respectively.

Part b· 7 steps

Use the matrices in (a) above to determine the total expenditure incurred by each trader and hence determine the maximum difference between the traders.

  1. Calculate the total expenditure for each trader by multiplying the quantity matrix by the price matrix.

    Expenditure Matrix (E)=Q×P=[362045183825][170210]\text{Expenditure Matrix (E)} = Q \times P = \begin{bmatrix} 36 & 20 \\ 45 & 18 \\ 38 & 25 \end{bmatrix} \begin{bmatrix} 170 \\ 210 \end{bmatrix}

  2. Perform the matrix multiplication.

    For Juma: (36×170)+(20×210)=6120+4200=10320(36 \times 170) + (20 \times 210) = 6120 + 4200 = 10320 Ksh
    For Tom: (45×170)+(18×210)=7650+3780=11430(45 \times 170) + (18 \times 210) = 7650 + 3780 = 11430 Ksh
    For Mary: (38×170)+(25×210)=6460+5250=11710(38 \times 170) + (25 \times 210) = 6460 + 5250 = 11710 Ksh

    E=[103201143011710]E = \begin{bmatrix} 10320 \\ 11430 \\ 11710 \end{bmatrix}

    The total expenditure incurred by each trader is:
    Juma: Ksh 10320
    Tom: Ksh 11430
    Mary: Ksh 11710

  3. Determine the maximum difference between the traders.

    Difference between Tom and Juma = 11430−10320=111011430 - 10320 = 1110 Ksh
    Difference between Mary and Juma = 11710−10320=139011710 - 10320 = 1390 Ksh
    Difference between Mary and Tom = 11710−11430=28011710 - 11430 = 280 Ksh

    The maximum difference between the traders is Ksh 1390\boxed{\text{Ksh 1390}}.

    c) Later in the week the cost of maize flour decreased by 10% while the price of sugar increased by 20%. Juma and Mary bought the same number of items as they did earlier in the week. Use matrix method to find the expenditure of each trader.

  4. Calculate the new prices.

    New price of maize flour = 170×(1−0.10)=170×0.90=153170 \times (1 - 0.10) = 170 \times 0.90 = 153 Ksh
    New price of sugar = 210×(1+0.20)=210×1.20=252210 \times (1 + 0.20) = 210 \times 1.20 = 252 Ksh

  5. Create the new price matrix.

    New Price Matrix (Pnew)=[153252]\text{New Price Matrix (P}_{\text{new}}\text{)} = \begin{bmatrix} 153 \\ 252 \end{bmatrix}

  6. Calculate the new expenditure for each trader using the original quantity matrix and the new price matrix.

    New Expenditure Matrix (Enew)=Q×Pnew=[362045183825][153252]\text{New Expenditure Matrix (E}_{\text{new}}\text{)} = Q \times P_{\text{new}} = \begin{bmatrix} 36 & 20 \\ 45 & 18 \\ 38 & 25 \end{bmatrix} \begin{bmatrix} 153 \\ 252 \end{bmatrix}

  7. Perform the matrix multiplication.

    For Juma: (36×153)+(20×252)=5508+5040=10548(36 \times 153) + (20 \times 252) = 5508 + 5040 = 10548 Ksh
    For Tom: (45×153)+(18×252)=6885+4536=11421(45 \times 153) + (18 \times 252) = 6885 + 4536 = 11421 Ksh
    For Mary: (38×153)+(25×252)=5814+6300=12114(38 \times 153) + (25 \times 252) = 5814 + 6300 = 12114 Ksh

    The expenditure of each trader with the new prices is:
    Juma: Ksh 10548\boxed{\text{Ksh 10548}}
    Tom: Ksh 11421\boxed{\text{Ksh 11421}}
    Mary: Ksh 12114\boxed{\text{Ksh 12114}}

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